CAIE S2 2012 November — Question 4 7 marks

Exam BoardCAIE
ModuleS2 (Statistics 2)
Year2012
SessionNovember
Marks7
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TopicLinear combinations of normal random variables
TypeSame variable, two observations
DifficultyChallenging +1.8 This question requires students to recognize that they need to work with the difference of random variables (specifically 2X - Y where X and Y are independent normal distributions), then find P(X ≥ 2Y) = P(X - 2Y ≥ 0). This involves understanding linear combinations of normal variables, calculating the mean and variance of the combined distribution, and standardizing. The conceptual leap to reformulate the problem and handle the variance of 2Y correctly (variance becomes 4×1550) makes this significantly harder than routine S2 questions, though it's still within the syllabus scope.
Spec2.04e Normal distribution: as model N(mu, sigma^2)5.04b Linear combinations: of normal distributions
4 The masses of a certain variety of potato are normally distributed with mean 180 g and variance \(1550 \mathrm {~g} ^ { 2 }\). Two potatoes of this variety are chosen at random. Find the probability that the mass of one of these potatoes is at least twice the mass of the other.
Question 4:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Use of \(X_1 - 2X_2\) or similar; \(E(X_1 - 2X_2) = 180 - 360 = -180\)B1 Or use of \(\frac{1}{2}X_1 - X_2\); \(E(2X_1-X_2)=360-180=180\); or \(E(\frac{1}{2}X_1-X_2)=90-180=-90\)
\(\text{Var}(X_1 - 2X_2) = 5 \times 1550\) or \(7750\)M1 A1 for \(1550 + 4\times 1550\) or \(\frac{1}{4}\times 1550 + 1550\); \(7750\) or \(1937.5\)
\(\dfrac{0-(-180)}{\sqrt{7750}}\) or \(\dfrac{0-180}{\sqrt{7750}} = \pm 2.045\)M1 Allow incorrect var (dep \(> 0\) & \(\neq 1550\)), no follow-through; standardising – no mixed methods; or \(\pm(0--90)/\sqrt{1937.5}\)
\(1 - \Phi(2.045) = 0.0205\) or \(0.0204\)M1 A1 For finding correct area (consistent with working)
Ans \(0.041\) (2 sf)B1ft [7] Allow double their prob 
## Question 4:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of $X_1 - 2X_2$ or similar; $E(X_1 - 2X_2) = 180 - 360 = -180$ | B1 | Or use of $\frac{1}{2}X_1 - X_2$; $E(2X_1-X_2)=360-180=180$; or $E(\frac{1}{2}X_1-X_2)=90-180=-90$ |
| $\text{Var}(X_1 - 2X_2) = 5 \times 1550$ or $7750$ | M1 A1 | for $1550 + 4\times 1550$ or $\frac{1}{4}\times 1550 + 1550$; $7750$ or $1937.5$ |
| $\dfrac{0-(-180)}{\sqrt{7750}}$ or $\dfrac{0-180}{\sqrt{7750}} = \pm 2.045$ | M1 | Allow incorrect var (dep $> 0$ & $\neq 1550$), no follow-through; standardising – no mixed methods; or $\pm(0--90)/\sqrt{1937.5}$ |
| $1 - \Phi(2.045) = 0.0205$ or $0.0204$ | M1 A1 | For finding correct area (consistent with working) |
| Ans $0.041$ (2 sf) | B1ft [7] | Allow double their prob |

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4 The masses of a certain variety of potato are normally distributed with mean 180 g and variance $1550 \mathrm {~g} ^ { 2 }$. Two potatoes of this variety are chosen at random. Find the probability that the mass of one of these potatoes is at least twice the mass of the other.

\hfill \mbox{\textit{CAIE S2 2012 Q4 [7]}}
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